In the figure shown, what is the current (in ampere) drawn from the battery? You are given R1 = 15 o

Q: Why are R3, R4, and R5 in series? They look like they are on different sides of a square!

In circuit analysis, physical shape doesn't matter. What matters is the path of the current. Since there are no junctions between R3, R4, and R5 where current could split, the exact same current must flow through all three. Therefore, they are in series.

Q: Why isn't R1 in series with R2? They are right next to each other.

Look at the point where R1 and R2 meet. There is a third wire connecting to R3. This is a junction. The current flowing through R1 will split at this junction, so R1 and R2 do not share the same current. Hence, they are not in series.

Q: How do I know when to use the parallel formula?

Two components are in parallel if they are connected across the exact same two nodes. In our circuit, after combining R3, R4, and R5 into Rs, both Rs and R2 are connected between the top junction and the bottom junction. This guarantees they have the same potential difference.

Q: Can I use Kirchhoff's laws instead of simplifying the circuit?

Yes, you absolutely can! You could set up loop equations and solve for the currents. However, for a circuit with a single voltage source and a reducible resistor network, finding the equivalent resistance is usually much faster and less prone to algebraic errors.

Q: What if the battery had an internal resistance?

If the battery had an internal resistance 'r', you would simply treat it as another resistor in series with the main circuit. You would add 'r' to your total equivalent resistance Req in the final step before calculating the current.

Question

In the figure shown, what is the current (in ampere) drawn from the battery? You are given $R_1 = 15 \Omega, R_2 = 10 \Omega, R_3 = 20 \Omega, R_4 = 5 \Omega, R_5 = 25 \Omega, R_6 = 30 \Omega, E = 15 	ext{ V}$

Options:

(A) 13/24

(B) 7/18

(C) 20/3

(D) 9/32

Sigma Percentile · Accepted Answer

### Decoding the Circuit Maze

Imagine you are an electron standing at the positive terminal of a $15	ext{ V}$ battery. You look ahead and see a complex maze of six different resistors.

Your mission is to find out the total current flowing out of the battery. To do this, we cannot just guess; we need to systematically break down this maze into a single, equivalent path.

We will use the powerful tools of series and parallel combinations to simplify the circuit step by step.

### The Right Flank: A Simple Series

Let's start our attack from the side furthest from the battery. Look closely at the right flank of the circuit, where resistors $R_3$, $R_4$, and $R_5$ are located.

Notice how they are connected end-to-end like a chain. There are no junctions or alternate paths between them. This is the classic definition of a **series combination**.

In a series circuit, the equivalent resistance is simply the sum of the individual resistances. So, we can replace these three resistors with a single equivalent resistor, $R_s$.

$R_s = R_3 + R_4 + R_5$

Substituting the given values, we get:

$R_s = 20 + 5 + 25 = 50\,\Omega$

### The Middle Ground: A Parallel Path

Now, our circuit has shrunk. We have replaced the entire right loop with a single $50\,\Omega$ resistor.

Look at this new resistor $R_s$ and the middle resistor $R_2$. They are connected across the exact same two nodes. This means they share the same potential difference, making them a **parallel combination**.

To find their equivalent resistance, $R_p$, we use the product-over-sum rule for two parallel resistors.

$R_p = \frac{R_2 	imes R_s}{R_2 + R_s}$

Plugging in our numbers:

$R_p = \frac{10 	imes 50}{10 + 50} = \frac{500}{60} = \frac{25}{3}\,\Omega$

### The Final Loop: Bringing It All Together

The maze is almost completely solved! We have reduced the entire right and middle sections into a single resistor $R_p$.

Now, look at the remaining circuit. We have $R_1$, our new $R_p$, and $R_6$ connected in a single, continuous loop with the battery.

Since the same current must flow through all of them, they are in **series**. We can find the total equivalent resistance, $R_{eq}$, by adding them up.

$R_{eq} = R_1 + R_p + R_6$

$R_{eq} = 15 + \frac{25}{3} + 30$

$R_{eq} = 45 + \frac{25}{3} = \frac{135 + 25}{3} = \frac{160}{3}\,\Omega$

### Ohm's Law to the Rescue

We have successfully reduced the entire complex network into a single equivalent resistance of $\frac{160}{3}\,\Omega$.

Now, we can finally answer the main question: what is the total current drawn from the battery? We call upon **Ohm's Law**, which states that current equals voltage divided by resistance.

$I = \frac{E}{R_{eq}}$

Substituting our total voltage and total resistance:

$I = \frac{15}{160/3} = \frac{45}{160}$

Simplifying the fraction by dividing the numerator and denominator by 5, we get our final answer.

$I = \frac{9}{32}	ext{ A}$

And just like that, by breaking the problem down into manageable atomic steps, we have conquered the circuit!

Animated Solution for Physics - Current Electricity: In the figure shown, what is the current (in ampere) drawn from the battery? You are given $R_{1} = 15Ω, R_{2} = 10Ω, R_{3} = 20Ω, R_{4} = 5Ω, R_{5} = 25Ω, R_{6} = 30Ω, E = 15 V$

Select Answer:

Visualized Solution

Circuit Analysis

Identifying Series Combination

Calculating $R_{s}$

Identifying Parallel Combination

Calculating $R_{p}$

Final Series Combination

Calculating $R_{e q}$

Applying Ohm's Law

Food for Thought

The Sigma Insight: Combination of Resistors

Decoding the Circuit Maze

The Right Flank

The Middle Ground

The Final Loop

Ohm's Law to the Rescue

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